However I can't seem to get anything close to what the questions asking me as when I tried solving that set equations I kept on getting equations that just cancelled out to 0 as both the sides worked out to be the same?

You can assume a < b < c. It follows that b+c > a+b > a+c, so you know what order the second sequence will be in.
Try it from the other end. Look at differences between consecutive terms of the second sequence, and take the difference of the differences.

However I can't seem to get anything close to what the questions asking me as when I tried solving that set equations I kept on getting equations that just cancelled out to 0 as both the sides worked out to be the same?

I would mess around with ##\displaystyle \ \frac{1}{b+c} , \frac{1}{c+a} , \frac{1}{a+b} \ ## just to see if I could find some connection with the squares of a, b, and c .

For [itex]\frac{1}{b+c}[/itex], [itex]\frac{1}{a+c}[/itex], [itex]\frac{1}{a+b}[/itex] to make arithmetic progression it shall be enough to check if [itex]\frac{1}{a+c}[/itex] - [itex]\frac{1}{b+c}[/itex] = [itex]\frac{1}{a+b}[/itex] - [itex]\frac{1}{a+c}[/itex]. Thanks for the ordering already established by previous posts.

And that easily simplifies to (b-a)(a+b)(a+c) = (c-b)(a+c)(b+c). That in turn is b2 - a2 = c2 - b2, which was given.